RVHS Prelim P2 Soln

1 River Valley High School 9729/02 [Turn over 2018 Preliminary Examinations MARK SCHEME 1 (a) W: Si SiCl4 + 2H2O → SiO2 + 4HCl X: P PCl5 + CH3COOH → POCl3 + CH3COCl OR 3CH3COOH + PCl3 ⎯→ 3CH3COCl + H3PO3 [4] (b) (i) Al2O3(s) + 2NaOH(aq) → 2Na[Al(OH)4](aq) + H2O(l) OR Al2O3(s) + 2OH−(aq) → 2[Al(OH)4]−(aq) + H2O(l) [3] (b) (ii) On addition of HC l(aq), a white ppt is formed. Ppt would dissolve in excess HCl(aq) to give a colourless solution. (c) BaCO3 has a higher thermal decomposition temperature. Ba2+ has a larger ionic radius compared to Z2+/Mg2+. Ba2+ hence has a lower charge density, and polarise the CO 32– anion to a lower extent (compared to Z2+/Mg2+), and the C–O bonds are weakened to a smaller extent, resulting in a higher thermal decomposition temperature. (d) The 3p electron to be removed from A l is at a higher energy level compared to the 3s electron to be removed fr om Mg, hence the p electron is less strongly attracted to the nucleus and require less energy to remove. [1] [Total: 9]

2 River Valley High School 9729/02 [Turn over 2018 Preliminary Examinations 2 (a) For pH = 3.5 [H+] = 10–3.5 = 3.16 x 10–4 moldm–3 Let the number of moles of C5H5NHCl be Y mol Since pKa = 5.25 , Ka = 10–5.25 = 5.62 × 10–6 (3.16 × 10–4)2 / Y – (3.16 × 10–4) = 5.62 × 10–6 Y = [(3.16 × 10–4)2 / 5.62 × 10–6 ] + (3.16 × 10–4) = 0.0181 mol Or (3.16 × 10–4)2 / Y = 5.62 × 10–6 Y = [(3.16 × 10–4)2 / 5.62 × 10–6 ] = 0.0178 mol Mass to be added = (0.0181 × (12.0 × 5 + 1.0 × 6 + 14.0 + 35.5)) ÷ 0.98 = 2.13 g (or 2.10 g) [3] (b) [C5H5N] = [(0.100 × 0.025) – (0.0125 × 0.005)] / 0.03 = 0.0813 mol dm–3 [pyridinium chloride] = (0.0125 × 0.005) / 0.03 = 2.08 × 10–3 mol dm–3 pKb =14 – pKa = 8.75 pOH = pKb + lg [salt]/[base] pOH = 8.75 + lg (2.08 × 10–3 / 0.0813) = 7.16 pH = 14 – 7.16 = 6.84 [3] (c) On addition of a small amount of acid (H +) to the buffer solution, nearly all the added H+ ions are neutralised by the large amount of C5H5N. Hence [H+] does not increase appreciably and the pH is kept approximately constant. C5H5N (aq) + H+(aq) → C5H5NH+(aq) On addition of a small amount of base (OH −) to the buffer solution, nearly all the added OH− ions are neutralised by the large amount of C5H5NH+. Hence [OH−] does not increase appreciably and the pH is kept approximately constant. C5H5NH+ (aq) + OH−(aq) → C5H5NH (aq) + H2O(l) [3] [Total: 9]

3 River Valley High School 9729/02 [Turn over 2018 Preliminary Examinations 3 (a) Initial P of NO(g) = ଵ ସ × 5 = 1.25 atm Initial P of NO2Cl(g) = 5 – 1.25 = 3.75 atm NO2Cl + NO ⇌ NOCl + NO2 Initial P / atm 3.75 1.25 0 0 Δ in P / atm –0.85 –0.85 +0.85 +0.85 Eqm P / atm 2.90 0.40 0.85 0.85 Kp = ሺ଴.଼ହሻሺ଴.଼ହሻ ሺଶ.ଽ଴ሻሺ଴.ସ଴ሻ = 0.623 (no units) [2] (b) [2] (c) [1] (d) Electrophilic addition [3] [Total: 8] PNO2Cl / atm time / min 0 t1 t2 t3 t4 2.90 3.25 3.75 O N Cl ×× ×× ×× •• •× •• ×× ×× ××

4 River Valley High School 9729/02 [Turn over 2018 Preliminary Examinations 4 (a) (i) Comparing Experiments 1 and 2, When the volume/concentration of HC l increases to 1.5 times, rate increases to 1.5 times. Hence, reaction is first order with respect to HCl. Comparing Experiments 1 and 3, When the volume/concentration of sucrose doubles, rate doubles. Hence, reaction is first order with respect to sucrose. [2] (ii) Rate = k [sucrose] [HCl] From Experiment 1, [sucrose] = 0.850 ൈ ଶ଴ ହ଴ = 0.340 mol dm−3 [HCl] = 1.23 ൈ ଷ଴ ହ଴ = 0.738 mol dm−3 k = ଵ.଻଻ ൈଵ଴షయ ሺ଴.ଷସ଴ሻሺ଴.଻ଷ଼ሻ = 7.05 ൈ 10−3 mol−1 dm3 min−1 [2] (iii) [sucrose] = 0.850 ൈ ସ଴ ଻଴ = 0.486 mol dm−3 [HCl] = 1.23 ൈ ଶ଴ ଻଴ = 0.351 mol dm−3 Both concentrations for Initial rate of reaction = 7.05 ൈ 10−3 ൈ 0.486 ൈ 0.351 = 1.20 ൈ 10−3 mol dm−3 min−1 [2] (b) (i) [1] 0

5 River Valley High School 9729/02 [Turn over 2018 Preliminary Examinations ( i i ) When [sucrose] is low, reaction is first order with respect to sucrose due to the availability of active sites on the enzyme molecules for binding. When [sucrose] is high, all active sites on the enzyme molecules are occupied. The rate of reaction t hen is independent of [sucrose] and the reaction is zero order with respect to sucrose. [2] (c) (i) [1] ( i i ) The zwitterions of aspartame form ion-dipole interactions with water. [1] ( i i i ) [3] [Total: 14]

6 River Valley High School 9729/02 [Turn over 2018 Preliminary Examinations 5 (a) Due to the similar energy/ close proxim ity of the 3d and 4s electrons in the transition elements, transtion element can form ions of approximately the similar stability by losing different number of electrons. [1] (b) In the presence of octahedral ligand fiel d, the degenerate d-orbitals in the metal complex were spilt into two energy levels. The colour observed is due to the difference in energy levels, ∆E. The difference in electronic configuration due to different oxidation state affect ∆E. Light of different energies/ different wavelengths are abs orbed for the promoion of electrons from the lower energy orbital to the va cant/partially filled higher energy orbital/ d-d transition, different complementary colour is observed. [2] (c) E⦵(Mn3+/Mn2+) is more positive than E ⦵([Mn(CN)6]3−/[Mn(CN)6]4−), which shows that Mn3+ is more readily reduced, henc e a weaker oxidising agent in cynaide solution than in water. More energy is required to add an electron to the negatively charged [Mn(CN)6]3− due to repulsion. [2] (d) (i) Figure 1.1 [1] ( i i ) Since [Mn(CN)6]3− ions are red, the energy gap between the 2 sets of d-orbitals in [Mn(CN) 6]3− is bigger. This suggests that [Mn(CN) 6]3− is the low spin complex, as its energy gap, ∆E, is greater than the pairing energy/ Coulombic repulsion / repulsion energy, electrons in [Mn(CN)6]3− would pair up. [1] (e) (i) [O] HMnO4− → MnO4− + H+ + e [R] 3H+ + 2e + HMnO4− → MnO2 + 2H2O Overall equation: 3HMnO4− + H+ → MnO2 + 2 MnO4− +2H2O [1] ( i i ) Ecell = +2.10 – (+0.90) = +1.20 V ∆G = −nFEcell = −(2)(96500)(1.20) = −232 kJ mol−1 [2] (f) In acidic condition, O2 + 4H+ + 4e ⇌ 2H2O E ⦵ = +1.23 V Mn3+ + e ⇌ Mn2+ E ⦵ = +1.54 V E⦵ cell = +1.23 – (+1.54) = −0.31 V [2] energy

7 River Valley High School 9729/02 [Turn over 2018 Preliminary Examinations In alkaline solution, O2 + 2H2O + 4e ⇌ 4OH− E ⦵ = +0.40 V E⦵ cell = +0.40 – (+0.18) = +0.22 V Since the E ⦵ cell > 0 in alkaline environment and E ⦵ cell < 0 in acidic environment, Mn( II) is more stable in acidic environment than in alkaline environment. [Total: 12]

8 River Valley High School 9729/02 [Turn over 2018 Preliminary Examinations 6 (a) (i) CO2: Carbon in CO2 has attained maximum oxidation state. N2: N≡N is very strong resulting high activation energy. [2] ( i i ) Methane: CH4(g) + 2O2 (g) → CO2(g) + 2H2O(l) Hydrogen sulfide: H2S (g) + 1½ O2 (g) → SO2(g) + H2O(l) Hydrogen: H2(g) + ½ O2 (g) → H2O(l) [2] ( i i i ) Energy evolved = 200 × 4.18 × (64.7 – 29.6) = 29.34 kJ mass of biogas used = 1000 × 6.44 × 10−4 = 0.644 g Fuel value of biogas = 29.34 ÷ 0.644 = 45.6 kJ g−1 [2] ( i v ) Amount of methane in 1 dm3 biogas = 0.644 × 0.722 ÷ (12.0 + 4.0) = 0.0291 mol Amount of H2S in 1 dm3 biogas = 0.644 × 0.012 ÷ (2.0 + 32.1) = 0.000227 mol Amount of H2 in 1 dm3 biogas = 0.644 × 0.027 ÷ 2.0 = 0.00869 mol Energy evolved = 29.34 kJ = (0.0291 × |ΔHc(CH4)|) + (0.000227 × 482) + (0.00869 × 386) ΔHc(CH4) = − 890 kJ mol−1 [2] (c) • Longer time for hot flue gas to pass through spiral copper coil • Spiral copper coil increase the surface area for energy transfer • copper is a good conductor of heat • the combustion takes place insi de the apparatus, not affected